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3 years ago

Mark can make 424242 birthday cakes in 777 days.

Mathematics

2 answers:

Zolol [24]3 years ago

8 0

Mark can make 303,030 birthday cakes.

ad-work [718]3 years ago

3 0

I think your quetsion was supposed to be
'Mark can make 42 birthday cakes in 7 days. How many birthday cakes can Mark make in 5 days?'

ratio and proprtion

cakes:days=42:7=42/7

42/7=x/5
42/7=x/5
6/1=x/5
times both sides by 5
30=x
30 cakes

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Simply the expression <br> 6(6+6+2v) HELP PLEASE ( I know my computer screen is dirty lol)

olya-2409 [2.1K]

Answer:

12v + 62

Step-by-step explanation:

distribute the 6 to 6, 6, and 2v.

36 + 36 + 12v=

12v + 72

8 0

3 years ago

) You are finally well!  But now your child is sick!  You know there have been mistakes in orders so you are on the internet t

rjkz [21]

Answer:

The doctor's dosage was not appropriate.

The right dosage is from 56.625 mg to 113.25 mg of Antibiotic to be given every 6 hour for the child with weight 25 lbs.

Step-by-step explanation:

It is given that 20 to 40 mg/kg/day is the recommended dosage.

Now, the doctor's order is 150 mg of antibiotic to be given every 6 hours. And my child weighs 25 lbs.

Now, 1 lb is equivalent to 0.453 kg.

So, my child's weight is (25 × 0.453) = 11.325 kg.

So, the doctor's order is 150 mg of antibiotic to be given every 6 hours for a child of weight 11.325 kg.

Hence, the dosage is [(150 × 4) ÷ 11.325] = 52.98 mg/kg/day.

So, this is not within the limit of 20 to 40 mg/kg/day.

Therefore, the doctor's dosage was not appropriate.

Now, let the appropriate dosage is x mg per every 6 hours for a child with weight 25 lbs i.e. 11.325 kg.

So, $20 \leq \frac{4x}{11.325} \leq 40$

⇒ $20\leq 0.3532x \leq 40$

⇒ $56.625 \leq x \leq 113.25$

So, the right dosage is from 56.625 mg to 113.25 mg of antibiotic to be given every 6 hours for the child with weight 25 lbs. (Answer)

5 0

3 years ago

Suppose the exponential regression equation for some data is y=3.7772•1.923^x. When x equals 5 what is the predicated value of y

sveticcg [70]

y = 3.7772. 1.923^5 = 99 to nearest whole number

3 0

3 years ago

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

aliina [53]

$y=x(x-2)^2$
$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
$\implies x=\dfrac23,x=2$

are the critical points, and judging by the picture alone, you must have $b=\dfrac23$ and $a=2$ . (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

$\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43$

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of $\dfrac{\mathrm dy}{\mathrm dx}=y'$ , which entails first finding the second derivative:

$y'=3x^2-8x+4$
$\implies y''=6x-8$

setting equal to 0 and finding the critical point:

$6x-8=0\implies x=\dfrac86=\dfrac43$

This is to say the minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ *occurs when $x=\dfrac43$ *, but this is not necessarily the same as saying that $\dfrac43$ is the actual minimum value.

The minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ is obtained by evaluating the derivative at this critical point:

$m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43$

4 0

3 years ago

HELP ASAP PLEASE!!! <br> 14 points!

Helga [31]

Your answer would be y=4x+17

8 0

3 years ago